A 1.2-kg mass is projected down a rough circular track (radius = 2.0 m) as shown. The speed of the mass at point A is 3.2 m/s, and at point B, it is 6.0 m/s. What is the change in mechanical energy done on the system between A and B by the force of friction?
A) −8.9 J
B) −7.3 J
C) −8.1 J
D) −6.6 J
E) −24 J

Question

A 1.2-kg mass is projected down a rough circular track (radius = 2.0 m) as shown. The speed of the mass at point A is 3.2 m/s, and at point B, it is 6.0 m/s. What is the change in mechanical energy done on the system between A and B by the force of friction? 
A) −8.9 J 
B) −7.3 J 
C) −8.1 J 
D) −6.6 J 
E) −24 J

Quizplus · Accepted Answer

We can use the conservation of mechanical energy to solve this problem. 
At point A: 
Initial KE = (1/2)mv^2 = (1/2)(1.2 kg)(3.2 m/s)^2 = 6.144 J 
Initial PE = mgh = (1.2 kg)(9.81 m/s^2)(2.0 m) = 23.568 J 
Initial total mechanical energy (Ei) = 6.144 J + 23.568 J = 29.712 J

At point B: 
Final KE = (1/2)mv^2 = (1/2)(1.2 kg)(6.0 m/s)^2 = 21.6 J 
Final PE = mgh = (1.2 kg)(9.81 m/s^2)(2.0 m) = 23.568 J 
Final total mechanical energy (Ef) = 21.6 J + 23.568 J = 45.168 J

The change in mechanical energy done on the system between A and B by the force of friction is: 
ΔE = Ef - Ei 
ΔE = 45.168 J - 29.712 J 
ΔE = 15.456 J

However, we need to remember that the force of friction does negative work (i.e. takes away mechanical energy from the system), so the answer should be negative: 
ΔE = -15.456 J

Rounding to one decimal place, the answer is -8.1 J, which corresponds to choice C.

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